Quote:
Originally Posted by Orgasmic Troll
if n = 999, then we have 99 0s from the ones digit and 90 0s from the tens digit
if n = 9999, we have 999 0s from the ones digit, 990 0s from the tens digit and 900 0s from the hundreds digit

except for the '0' we start with, the following code confirms your result & (corrected) formulae:
Code:
gp > ^Z
Suspended
(mfh@lx08 ~/NumberTheory/PARI) php
<?=($a=count_chars(join(range(0,99))))? $a[48]:0,"
",($a=count_chars(join(range(0,999))))? $a[48]:0,"
",($a=count_chars(join(range(0,9999))))? $a[48]:0,"
";^D
10
190
2890
(mfh@lx08 ~/NumberTheory/PARI) fg
gp
gp > calc0mfh(k)=k*10^(k1)(10^k1)/9+1
time = 0 ms.
gp > calc0mfh(3)
time = 0 ms.
%216 = 190
gp > calc0mfh(4)
time = 0 ms.
%217 = 2890
gp >
thus, the correct formula is:
number of '0's in {0,1,...,10^(k+1)1} = k*10^(k1)(10^k1)/9+1
I confirm your value for k=11 which is correct, i.e. seems to include the +1 already.
If your "iterations"(?) are ok, the result should be correct.